Hello Ankit,


>> Hi Manoj, you might want to check this method, because if you directly
> substitute Fx = df/dx = c(x,y)/a(x,y) and Fy = df/dy = c(x,y)/b(x,y), you
> get LHS = 2*c(x,y). You can take a look at
> http://geo.hmg.inpg.fr/loret/enseee/maths/enseee-maths-IBVPs-3.pdf ,
> which discusses the solution of the pdes of this type based on their
> classification.
>
>
> Thank you for pointing me to a source that, discusses the solution to
> pdes, based on their classification. However I'm afraid my question remains
> unanswered for the following reasons.
>


Firstly,  I think (I may be wrong) df / dx is equal to (dhof / dhox) (I
don't know how to write it here), only if f is a function of x, if not df =
(dhof / dhox) * dx + (dhof / dhoy) * dy , since f is a function of x and y,
so I suppose df / dx cannot be substituted back in the equation.


Secondly, I would like to cite this example in the research paper, by Dr.
Starrett( Aaron's professor indeed ) Lie
Groups<http://www.google.co.in/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CDEQFjAA&url=http%3A%2F%2Feuler.nmt.edu%2F~jstarret%2F05-649LieGroupODEFinalVersion.pdf&ei=WxdcUb65L4OErQeQz4HoAQ&usg=AFQjCNE9ugXh9NmZVLxq9LB8fpYuo6vusA&sig2=HLQVAGHySZ3PndKBIPECdA&bvm=bv.44697112,d.bmk>.
On page 16, he mentions clearly that to solve the equation, sx ξ + sy
η =
1.

It can be done by, ds = integral (dx / ξ) , I thought this might be for a
case where only ξ is a function of x, but he goes on to say on page 20 that
when ξ is equal to y, s would remain (x / y) . Sorry for being a bit
bookish here, but I hope my point is being put across.

It would be really helpful, if Aaron or other people here with a much
better mathematical background than me over here, could show me the way
ahead.
-- 
Regards,
Manoj Kumar,
Mech Undergrad.
BPGC
Blog <http://manojbits.wordpress.com>

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